AI 中文总结
本文通过关联秩不等式证明了四边均匀路径的线性Turán数猜想，给出边数上界$(r+1)n/r$，并刻画了等号条件为Steiner系统$S(2,r,r^2)$的并集。
AI 中文摘要
设$P_4^r$表示具有四条边的图路径的$r$-均匀扩张。我们证明每个$n$顶点线性$r$-均匀且不含$P_4^r$的超图至多有$(r+1)n/r$条边，从而解决了Adak和Verma对每个$r \geq 2$提出的猜想。等号恰好在顶点不相交的Steiner系统$S(2,r,r^2)$的并集时成立。主要工具是一个尖锐的关联秩不等式。若$N(H)$是线性$r$-均匀超图$H$的边-顶点关联矩阵，且其线图是余图（cograph），则$(r+1)\operatorname{rank}_{\mathbb{R}} N(H) \geq r|E(H)|$。等号成立当且仅当每个包含边的连通分量都是$S(2,r,r^2)$。证明遵循余图的并-连接分解。在连接节点处，各余子图的行差空间相互正交，可能的秩亏缺由平衡的余子图决定。Perron-Frobenius理论将最小的平衡部分识别为$r$个不相交的$r$-集合的平行类，而正交性论证将其数量限制为不超过$r+1$。等号情形随后重构出Steiner系统。线性Turán界由$\operatorname{rank}_{\mathbb{R}} N(H) \leq |V(H)|$得出。
英文摘要
Let $P_4^r$ denote the $r$-uniform expansion of the graph path with four edges. We prove that every $n$-vertex linear $r$-uniform $P_4^r$-free hypergraph has at most $(r+1)n/r$ edges, resolving a conjecture of Adak and Verma for every $r \geq 2$. Equality holds precisely for vertex-disjoint unions of Steiner systems $S(2,r,r^2)$. The main ingredient is a sharp incidence-rank inequality. If $N(H)$ is the edge-vertex incidence matrix of a linear $r$-uniform hypergraph whose line graph is a cograph, then $(r+1)\operatorname{rank}_{\mathbb{R}} N(H) \geq r|E(H)|$. Equality holds exactly when every edge-containing component is an $S(2,r,r^2)$. The proof follows the union-join decomposition of cographs. At a join node, the row-difference spaces of the co-components are mutually orthogonal, and the possible rank defect is determined by balanced co-components. Perron-Frobenius theory identifies the smallest balanced pieces as parallel classes of $r$ disjoint $r$-sets, while an orthogonality argument bounds their number by $r+1$. The equality case then reconstructs the Steiner system. The linear Turán bound follows from $\operatorname{rank}_{\mathbb{R}} N(H) \leq |V(H)|$.
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